The Unique MST（次小生成树）

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 22335		Accepted: 7922

Description

Given a connected undirected graph, tell if its minimum spanning tree is unique.

Definition 1 (Spanning Tree): Consider a connected, undirected graph G = (V, E). A spanning tree of G is a subgraph of G, say T = (V', E'), with the following properties:
1. V' = V.
2. T is connected and acyclic.

Definition 2 (Minimum Spanning Tree): Consider an edge-weighted, connected, undirected graph G = (V, E). The minimum spanning tree T = (V, E') of G is the spanning tree that has the smallest total cost. The total cost of T means the sum of the weights on all the edges in E'.

Input

The first line contains a single integer t (1 <= t <= 20), the number of test cases. Each case represents a graph. It begins with a line containing two integers n and m (1 <= n <= 100), the number of nodes and edges. Each of the following m lines contains a triple (xi, yi, wi), indicating that xi and yi are connected by an edge with weight = wi. For any two nodes, there is at most one edge connecting them.

Output

For each input, if the MST is unique, print the total cost of it, or otherwise print the string 'Not Unique!'.

Sample Input

2
3 3
1 2 1
2 3 2
3 1 3
4 4
1 2 2
2 3 2
3 4 2
4 1 2

Sample Output

3
Not Unique!

Source

POJ Monthly--2004.06.27 srbga@POJ

思路:求图的次小生成树，网上找的思路，先求mst，然后枚举删除mst中的每一条边，判断新得到的mst'是否等于mst

/*times memy
 63ms   804k
 by orchid
 */
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <iostream>
#define INF 1 << 25
using namespace std;
int n, m, ans;
int p[120], r[5000];
int u[5000], v[5000], w[5000];
bool flag[5000];//标记最小生成树中的所有边
int mst;//
void init()
{
    for(int i = 1; i <= n ; i++)
    p[i] = i;
    for(int i = 1; i <= m; ++i)
    r[i] = i;
}
int cmp(int i,int j)//按权值间接排序
{
    return w[i] < w[j];
}
int find(int x)//并查集路径压缩
{
    return p[x] == x? x:p[x] = find(p[x]);
}
void read_graph()
{
    cin >> n >> m;
    for(int i = 1; i <= m; ++i)
    cin >> u[i] >> v[i] >> w[i];
}
void kruskal()//求mst
{
    sort(r + 1,r + m + 1,cmp);
    for(int i = 1; i <= m; ++i)
    {
        int e = r[i];
        int x = find(u[e]);
        int y = find(v[e]);
        if(x != y){ ans += w[e];flag[e] = false; p[x] = y; }
    }
}
void kruskal2()//求次小生成树
{
    sort(r + 1,r + m + 1,cmp);
    for(int i = 1 ; i <= n; ++i) p[i] = i; //记得恢复并查集中的p[]
    for(int i = 1; i <= m; ++i)
    {
        int e = r[i];
        int x = find(u[e]);
        int y = find(v[e]);
        if(x != y){ ans += w[e]; p[x] = y; }
    }
}
void solve()
{
    int i;//初始化
    memset(flag,true,sizeof flag);
    read_graph();
    init();
         //求mst
    ans = mst  = 0;
    kruskal();
    mst = ans;

    int tp, tr;
    tp = tr = 0;
    for(i = 1; i <= m; ++i)
    if(!flag[i])//寻找mst中的边
    {
        ans = 0;
        w[tp] = tr;//恢复上一条去除的边

        tp = i;tr = w[i];
        w[i] = INF;//去除此条边

        kruskal2();
        if(ans == mst) break;
    }
     //cout << fst << ' ' << ans <<endl;
    if(i <= m) {cout<< "Not Unique!" <<endl; }
    else cout << mst <<endl;
}
int main()
{
    ios::sync_with_stdio(0);
    int t;
    cin >> t;
    while(t--)
    solve();
    return 0;
}